Dedekind Cuts - Constructing the Real Numbers (Part 4 - Step 4 #2) #4.3.1.4d

Constructing the real numbers with Dedekind Cuts. Here we finish step four with the hardest part, proving the additive inverse field axiom..

At last we are going to prove that real numbers exist! Using Dedekind Cuts. In this video we define an order on the set R of cuts. Then we prove that this set has the Least Upper Bound property.

After constructing the real number field, we will move on to the climax of this mini-series, proving the uniqueness of the real number field. We will then lay the foundation of multivariable calculus before moving on to complex numbers.

It turns out Elliptic Curves over Complex Numbers are central to Wiles's proof of Fermat's Last Theorem. With the first videos on this channel, we managed to keep the math very simple, almost exclusively nothing more advanced than algebra. Now, to complete our brief tour of elliptic curves, we have to take an ENORMOUS leap forward in the level of the mathematics involved.

I want to make sure we are all on the same page before we begin learning the more advanced stuff in earnest. I am going to assume we are all familiar with high school math and college math through first or second year calculus. I'm going to introduce complex numbers and complex integration in some detail with the next several videos. We need to examine multivariable calculus as well.

This series will culminate with the Weierstrass Equation, which is the thing that connects elliptic curves to complex numbers, and thus allows us to connect them to modular forms, which is what Wiles's proof is all about.

Here is the outline of this series on complex numbers:
0. Introduction
1. The Real Number System (We are here)
2. Complex Numbers
3. Complex Functions
4. Exponential and Trigonometric Functions
5. Complex Integration
6. Cauchy's Integral Theorem
7. Cauchy's Integral Formula
8. Laurent Series
9. Complex Residues
10. Lattices and Doubly Periodic Functions
11. Lattices and Tori and Groups
12. The Weierstrass p-Function
13. The Weierstrass Equation: Complex Functions and Elliptic Curves

Table of Contents This Video:
0:00 - Intro
0:25 - (A5) Additive Inverse
1:57 - The construction of β
6:25 - Proof β is a cut
6:37 - Property (I)
9:00 - Property (II)
11:29 - Property (III)
14:58 - Proof α + β = 0*
15:29 - Proof α + β ⊆ 0*
18:08 - Proof 0* ⊆ α + β

Please leave any questions, comments, or suggestions in the comments below!

Credits:
Music: "Oberkorn" and "St. Jarna" by Depeche Mode.
Bibliography: Principles of Mathematical Analysis, by Walter Rudin

[subhrajitdasgupta]

I have been following your series on Dedekind cuts. You try your best to make the students understand a topic as lucidly as possible. I was really finding it difficult while going through Rudin and looking for such a video. Thanks a lot for this video.

[greg]

A viewer named Karen asked me to explain again why w = -v/2. She seems to have deleted her question, but it's a great question (as all are), so I want to try to answer anyway.

Setting w = -v/2 is the crux of Rudin's proof. Remember beta is a cut, which means it has no largest number. No matter what element of β you choose, you can find one bigger. Rudin does this by finding a new number half way between the element and the upper limit of the cut.

What we're trying to do here is show that any element of 0* is also an element of α + β. So we start by picking an arbitrary element of 0*. That is, any v < 0 is an element of 0*. So we have to show that this number we chose randomly, v < 0, is also an element of α + β.

β is defined as all p such that -p - w ∉ α.  What in the world does this mean.  It means that for any element of β, the  of p is  in α.  But there is no largest element of β.  You can pick an element of β.  That means that the inverse of that slightly bigger element is still going to be outside α.  That's what -p - w ∉ α says.  It might be easier to rewrite it as –(p + w) ∉ α.  For any p ∈β, –(p + w) ∉ α.

So, for any element p of beta, we can pick a tiny w > 1 such that p + w is still in beta. Then, using the Archimedean property, we find n such that nw ∈ α,  but (n+1)w ∉ α and then also, (n + 2)w ∉ α.

So now, if our element of beta is the inverse of (n+2)w, then we're all set.  We'll be able to subtract w from the inverse of that element and still be outside α.  So we set p = -(n + 2)w.

Then, -p = (n + 2)w which is not in alpha, and -p -w = nw + 2w - w = nw - w = (n + 1)w, which is also not in alpha. So p is in β, because (n + 1)w ∉ α.

So, if p = -nw -2w, then p + nw = -nw -2w + nw = -2w. That means p + nw ∈ α + β.  And p + nw = -2w.  We have to show -2w is an element of 0* and we're done.  Okay, well, let's set w = -v/2, where v is an arbitrary element < 0.

Then, w = -v/2 which means 2w = -v which means v = -2w!  So we just showed that starting with any number less than 0, we can show that that number is also an element of α + β.

With proofs like this, the proof originally was almost invariably done in reverse--what do we need to end up with, then work backward to find something that will get us to that point. But then when the author presents the proof, they don't explain why they chose that number, they say "just watch" and it turns out how they want. But Rudin picked w = -v/2 because it leads to p + nw = -2w = v, which shows that an element of 0* is also an element of alpha plus beta.

[theunknownscientist]

19:37 Ok so, here is my objection against this, I've been struggling with this for 4 days now: from my understanding the archimedian proprety works with the existence of natural numbers, so for α<0*, it wouldn't work because w>0 and n≥0. So you would be stuck in getting nw>0 and not be able to touch negative numbers. But anyway, I think it can be proved that indeed it works with the integers. Even so, your α does not have a greatest element so if you would take say p∈α at random you could find(using the archimedian property) an integer n st wn>p. But for p, you know there is a q∈α st p<q. Soo yeah, wn would be in α if wn<q. But how can you prove that? as a matter of fact, if you go on, and find more r∈α, st wn<r, how can you be so sure that, there isn't a t∈α, st (n+1)w<t? I mean graphically yes, you are right, but the graphic representation of the reals is dependent on how you define them, so the archimedian property partitions the reals, if and only if they you already constructed them(i.e. the construction is done prior to proving the archimedian property for the reals, even if there it applies for the rationals). Is there some kind of theorem or lemma that says that wn∈α for n being the largest st w(n+1) would not be in α? If so, why didn't you stated it here? Even more, the archimedian property does not appear in such a form anywhere, not even in Rudin's book, I miss to see where is the rigour in any of the proofs.

[congo]

Hi Greg,
I couldn't make the connection between the archemedian property and the existence of an n such that nw is in alpha but (n+1)w is not in alpha. Could you please elaborate on that. I'm talking with reference to the proof of additive inverse that 0* is a subset of alpha+alpha*.
Thanks

[zcahhf]

[8:36] Why -q ∉ 𝛽? Graphically it makes sense, but analytically? I'm confused. Thank you.

[Kanrop.h]

20:15 what if \alpha < 0* the aggrument that there is n such that nw in alpha but (n+1)w is not in alpha still true ?